[英]how to pass a values from the database to the json in php
在此先感谢,我在表列中有三个薪水数据。 我正在使用while循环在视图页面中显示三个薪水值。 但是我必须将这三个值传递给json中的三个变量,例如{salary1:$ sal1,salary2:$ sal2,salary3:$ sal3}。 如何将循环的三个薪水值分成三个变量
My code as below:
<table border="1">
<caption><h2>View Registration</h2></caption>
<tr>
<th>Name</th>
<th>Designation</th>
<th>Email</th>
<th>Salary</th>
<tr>
<?php
include('common.php');
$sql = mysql_query("select * from register");
while($row = mysql_fetch_array($sql))
{
?>
<tr>
<td><?php echo $row['name']?></td>
<td><?php echo $row['designation']?></td>
<td><?php echo $row['email']?></td>
<td><?php
$salary = $row['salary'];
echo $salary;
?></td>
<tr>
<?php
}
?>
</table>
创建一个包含所有薪水并使用json_encode回显的数组,该数组将根据需要转换为json,
$sql = mysql_query("select * from register");
$salary =array();
while ($row = mysql_fetch_array($sql)) {
$salary[$row['name']] = $row['salary'];
}
echo json_encode($salary);
<table border="1">
<caption><h2>View Registration</h2></caption>
<tr>
<th>Name</th>
<th>Designation</th>
<th>Email</th>
<th>Salary</th>
<tr>
<?php
include('common.php');
$sql = mysql_query("select * from register");
$arraySalary=array();// declaring an array for json array
$loopCounter=1;
while($row = mysql_fetch_array($sql))
{
?>
<tr>
<td><?php echo $row['name']?></td>
<td><?php echo $row['designation']?></td>
<td><?php echo $row['email']?></td>
<td><?php
$salary = $row['salary'];
$arraySalary['salary'.$loopCounter]= $row['salary']; // making an array os diffrent index as salry1, salary2, salary3.
$loopCounter++;
?></td>
<tr>
<?php
}
echo json_encode($arraySalary); // here you will get the salary json.
?>
</table>
请尝试一下。 我已经检查过了。 并且工作正常。
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